Recent content by ChrisJ

Yes, thank you!

Sorry! Yes, I just saw it, the ##\partial_c## bit, Thank you! I just saw it, ok, so the terms with ##\Gamma^u_{\theta\theta}## and ##\Gamma^u_{\phi\phi}## lead to ##\frac{-4}{v-u}\frac{\partial f}{\partial u}## (and the same for the ones with ##\Gamma^v_{\theta\theta}## and...

Yes, but only for ##\Gamma^u_{\theta\theta}## and ##\Gamma^u_{\phi\phi} ## (and the same, but with with ##v## as ##c##). But that would lead to partials wrt to ##\theta## and ##\phi##, which don't appear in what I am trying to reproduce.

Previous to trying this I found all of the christoffel symbols for the metric define in the OP, so now trying with this new definition, it still simplifies to ##g^{ab}\partial_a \partial_b ## because ##a## and ##b## can only take on ##u## or ##v## since there are not partials wrt to the others...

Oh. Ok.. thanks will give that a go!

Was not sure weather to post, this here or in differential geometry, but is related to a GR course, so... I am having some trouble reproducing a result, I think it is mainly down to being very new to tensor notation and operations. But, given the metric ##ds^2 = -dudv + \frac{(v-u)^2}{4}...

Oh right ok, there might be an error there somewhere but that is a wrong method anyway. As per post 6 onwards tried to go about the problem from a different angle as per Orodruin's suggestion. However, I am pretty certain that ##\frac{d\theta}{ds} = \sin \theta ##, which that bit is still needed...

Where did you get that from? From post 5, it can be seen that ds^2 = \csc^2 \theta d\theta^2 \\ ds = \csc \theta d \theta \\ ds = \frac{d\theta}{\sin \theta}

Ok, I think I am having a bit of trouble here, does not seem to equal zero the way I'm doing it. I found all the derivatives I need separately to make it clearer, then plug in at the end. \frac{d}{ds} r \sin \theta = r \cos \theta \frac{d\theta}{ds} = r \cos \theta \sin \theta \\...

ok its been a while since I've done implicit differentiation, but am I correct in thinking that, for example ##\frac{d}{ds} r \sin \theta = r \cos \theta \frac{d\theta}{ds} = r \cos \theta \sin \theta ## Since in my wrong method above I found that ##\frac{d\theta}{ds}= \sin \theta ##

Oh, bugger. Ok thanks. Well the geodesic equation, given the chrsitoffel symbols to the original coordinate system is as follows 0 = \frac{d^2x}{ds^2} - \frac{2}{y}\frac{dx}{ds}\frac{dy}{ds} \\ 0 = \frac{d^2y}{ds^2} - \frac{1}{y} \left( \frac{dy}{ds} \right)^2 + \frac{1}{y}\left(...

The idea was to find ##ds^2=##...##d\theta^2## to then find the non-vansihing christoffel symbols of metric in terms of ##\theta## to plug into new geodesic equation. This is what I have done so far now, leaving where last post left off and fixing mistake ds^2 = \left( \frac{1}{r^2 \sin^2...

Sorry, I cannot see that? What I did was this, ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( -r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\ ds^2 = \frac{-r^2 \sin^2 \theta d\theta^2}{r^2 \sin^2 \theta} + \frac{r^2 \cos^2 \theta d \theta^2}{r^2 \sin^2...

Given ##ds^2 = y^{-2}(dx^2 + dy^2)##, I am trying to prove that a demicircle centred on the x-axis, written parametrically as ##x=r\cos\theta + x_0 ## and ##y= r \sin \theta ## are geodesics. Where ##r## is constant and ##\theta \in (0,\pi)## I have already found the general form of the...

Ok thanks both, So, if I am understanding you both correct, something like ##ds^2 = -xdv^2 + 2dvdx## would be g = \begin{pmatrix} -x & 1 \\ 1 & 0 \end{pmatrix}